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Current time:0:00Total duration:24:15

Video transcript

the whole point in learning differential equations is that eventually we want to model real physical systems I know everything we've done so far has really just been a toolkit of being able to solve them but the whole reason is that is that because differential equations can describe a lot of systems and then we can actually model them that way and we know that in the real world everything isn't these nice continuous functions so over the next couple of videos we're going to talk about functions that are a little bit more discontinuous than what you might be used to even in kind of a traditional calculus or a traditional precalculus class and the first one is the unit step function let's write it is U and they'll put a little subscript C here of T and it's defined as when T is let me put it this way it's defined as 0 when T is less than whatever our subscript I put here when T is less than C and it's defined as 1 that's why we call it the unit step function when T is greater than or equal to C and if I had to graph this you could graph it as well but it's not too difficult to graph let me draw my x-axis right here and I'll do a little thicker line that's my x-axis right there this is my y-axis y-axis right there and let's and when we talk about Laplace transforms which we'll talk about shortly we only care about T is greater than 0 because we saw our definition of the Laplace transform we're always taking the integral from 0 to infinity so we're only dealing with the positive x-axis but anyway if by this definition it would be 0 all the way until you get to some value C so you'd be 0 until you get to C and then at sea you jump and the Point C is included X is equal to C here so it's included so I'll put a dot there because it's greater than or equal to C you're at 1 so this is 1 right here and then you go forward for all of time and you're like Sal you just said that differential equations we're going to model things why is this type of a function useful well in the real world sometimes you do have something that essentially jolts something that mu from this position to that position and obviously nothing can move it immediately like this but you might have some some system you know it could be an electrical system or mechanical system where where maybe the behavior looks something like this where it's you know maybe it moves it like that or something and this function is a pretty good analytic approximation for some type of real-world behavior like this when something just gets moved whenever we solve these differential equations analytically we're really just trying to get a pure model of something eventually we'll see that it you know doesn't perfectly describe things but it helps describe it enough for us to get a sense of what's what's going to happen sometimes it will completely describe things but anyway let's we can ignore that for now so let me get rid of these things right there so the the first question is well you know what if something doesn't jar just like that what if what if I want to construct more fancy unit functions or or more fancy step functions let's say I wanted to construct something that looked like this let me say this is my this is my y-axis this is my x-axis and let's say I wanted to construct something that is at let's say it's that let me do it a different color let's say it's at 2:00 until I get to I don't know let's say it's a 2 until I get to PI until I get to PI and then from PI until forever it just stays it it just stays at 0 so how could i how could I construct this function right here using my unit step function so what if let's see what if I had written it as so my unit step function is 0 initially so what if I make it to minus a unit step function let me say to minus a unit step function front that starts at PI so if I define my function here will this work well this unit step function when we pass PI is only going to be equal to 1 but we want this thing to equal to 0 so it has to be 2 minus 2 so I'll have to put a 2 here and this should work when we go from when we're at any value below pi T is less than PI here this becomes a zero so our function will just evaluate to two which is right there but as soon as we hit T is equal to PI that pi is the C in this example as soon as we hit that the unit step function becomes one we multiply that by two and we have two minus two and then we end up here with zero now that might be nice and everything but let's say you wanted to let's say you wanted it for it to to go back up let's say that instead of instead of it going like this instead of it going like this let me kind of erase that by over drawing the x-axis again we want the function to jump up again we want it to jump up again at let's say it's some value let's say it's at 2 pi we want the function to jump up again how could could we construct this and let's and we could make it jump to anything let's say we want it to jump back to two well we could just add another unit step function here something that would have been zero all along all the way up till this point but then at 2 pi it jumps so that it would be in this case our C would be 2 pi so unit step function we wanted to jump to 2 this would just jump to 1 by itself so let's multiply it by 2 and now we have this function so you could imagine you can make an arbitrarily complicated function of things jumping up and down to different levels based on different essentially linear combinations of these unit step functions now what if I wanted to do something a little bit fancier what if I wanted to do something that let's say I have some function let's say have some function that looks like this let me draw some function that's I should draw straighter than that I should have some standards so let's say that just my regular f of t f of T let me this is X actually why am i doing X this would be the T axis we're dealing the time domain it could have been X but and then this is call this f of T so let me draw some arbitrary f of T let's say my function looks something crazy like that now what if I modeling a physical so this is my F of T what if I'm modeling a physical system that doesn't do this that actually at some point well I say let's say it stays at zero it stays at zero for until some value let's say this goes to zero until I'll call that C again and then at C F of T kind of starts up so right at C f of T should start up so it just it kind of goes like this so essentially what we have here is a combination of it's zero all the way and then we have a shifted f of T so at C we have a shift at F of T so it shifts that way so how can we construct this yellow function where it's essentially a shifted version of the screen function but it's zero below C this green function might have continued it might have gone something like this it might have you know continued and done something crazy but what we did is we shifted it from here to there and then we zeroed out everything before C so how could we do that well just shifting this function you've learned in your algebra 2 or your precalculus classes to shift a function by C to the right you just replace your T with a t minus C so this function right here is f of T minus C and to make sure I get it right what I always do is I imagine okay what's going to happen when T is equal to C when T is equal to C you're going to have a C minus a C and you're going to have F of 0 so f of 0 is should be the same so when T is equal to C this value the value of the function should be equivalent to the value of the original green function at 0 so it's equivalent to that function value which makes sense if we go up one more above C so let's say we let's say this is one more above C so we get to this point when you if this if T is C plus 1 then when you put C plus 1 minus C you just have F of 1 and F of 1 is really just this point right here and so it'll be that F of 1 so it makes sense so as we move one forward here we're essentially at the same function value as we were there so the shift works but I said that we had to also if I just shifted this function you would have all this other stuff because you would have had all this other stuff when the function was back here still going on right the function I'll draw it lightly it would still continue but I said I wanted to zero out this function before we reach C so how can I zero out that function well it's I think it's pretty obvious to you I started this video talking about the unit step function so what if I multiply the unit step function times this thing what's going to happen so what if I my new function I call it the unit step function up till C of T times f of T minus C so what's going to happen this until we get to C right the unit step function is zero when it's less than C so you have zero times I don't care what this is zero times anything is zero so this function is going to be zero once you hit C the unit step function becomes one so once you pass C this thing becomes a 1 and you're just left with 1 times your function so then your function can behave is it would like to behave and you actually shifted it this t minus C is what actually shifted the screen function over to the right and this is actually going to be a very useful way to look at or very usefully constructed function and in a second we're going to figure out the Laplace transform of this and you're going to appreciate I think why this is a useful function to look at but now you understand at least what it is and and and and why it essentially shifts a function and zeroes out everything before that point well I told you that this is a useful function so it I we should add its Laplace transform to our library of Laplace transforms so let's do that so let's take the Laplace transform let's take the Laplace transform of this of the unit step function up to C I'm doing it in fairly general terms in the next video we'll do a bunch of examples when we can apply this but we should at least prove to ourselves what the Laplace transform of this thing is well the Laplace transform of anything it's or our definition so far is the integral from 0 to infinity of e to the minus s T times our function so our function in this case is the unit step function U sub C of T times f of t minus C DT and it seems very general it seems very hard to evaluate this integral at first but maybe we can make some form of a substitution to get into a term that we can appreciate so let's make a substitution here let me pick a nice a nice variable to work with I don't know we're not using an X anywhere which might as well use an X that's the most fun variable to work with X sometimes a lot of you'll see in a lot of math classes they introduce these crazy Latin alphabets and that by itself makes it hard to understand but so I like to stay away from those crazy Latin alphabets so we'll just we'll just use a regular ax let's make a substitution let's say that X is equal to t minus C right or you could if we added C to both sides we could say that T is equal to X plus C let's see what happens to our substitution and also if we took the derivative of this both sides of this which I or again just a differential you would get DX is equal to DT I mean if you took DX with respect to DT you would get that if T equal to 1 C is just a constant and if you multiplied both sides by DT you get DX is equal to DT and that's a nice substitution so what is our integral going to become with this substitution so our integral it's this was T equals 0 to T is equal to infinity when T is equal to 0 when T is equal to 0 what is X going to be equal to well X is going to be equal to minus C actually before I go there let me actually take a step back because I we could progress let me we could go in this direction but we could actually simplify it more before we do that let's go back to our original integral before we even made our substitution if we're taking the integral from 0 to infinity of this thing we already what does this integral look like what does this function look like when it's zero we have we have this unit step function sitting right here we have the unit step function sitting right there so this whole expression is going to be zero until we get to C right this whole thing by definition this unit step function is zero until we get to see so this everything's going to be zeroed out until we get to C so we could we could essentially say you know we don't have to take the integral from T equals zero to T equals infinity we could take the integral let me write it here we could take the let me I'll just use that old integral sign we could just take the integral from T is equal to C 2 T is equal to infinity of e to the minus st the unit step function you see if T times f of t minus C DT in fact at this point this unit step function is kind of it doesn't it has no use anymore because before before T is equal to C it's zero and now that we're only worried about values above C it's equal to 1 so it equals 1 in this context I want to make that very clear to you what did I do just here I changed our bottom boundary from 0 to C and I think you might realize why I did it when I was working with the substitution and because this will simplify things if we do this ahead of time so for if we have this unit step function this thing is going to zero out this entire integral before we get to C remember this definite integral is really just the area under this curve right of this whole function of the unit step function times all of this stuff all of this stuff when we multiply it it's going to be 0 until we get to some value C and then above C it's going to be e to the minus st times F of t minus C so it's going to start doing all this crazy stuff so if we want to take the the essentially find the area under this curve we can we can ignore all the stuff that happens before C so instead of going from T equals 0 to infinity we can go from T is equal to C to infinity because there is no area before T was equal to C so that's all I did here and then the other thing I said is that the unit step function it's going to be won over this entire over this entire range of potential T values so we can just kind of ignore it right it's just going to be one this entire time so our integral simplifies to the definite integral from T is equal to C 2 T is equal to infinity of e to the minus st times F of t minus C DT and this will simplify it a good bit I was going down the other Road when I was in the substitution first which would have worked but I think the argument as to why I could have changed the boundaries would have been a harder argument to make so now that we had this let's let's go back and make that substitution that X is equal to t minus C so our integral becomes I'll do it in green when T is equal to C what is X then X is 0 right C minus C is 0 when T is equal to infinity what is X well X is you know infinity minus any constant still going to be an infinity or the limit as T approaches infinity X is still going to be infinity here time and this the integral of e to the minus s but now instead of a T we have the substitution if we said X is equal to t minus C then we can just add C to both sides you get T is equal to X plus C so you get X plus C there and then times the function f of t minus C but we said t minus C is same thing as X and DT is the same thing as DX showed you that right there so we can write this as DX this is starting to look a little bit interesting so what is this equal to this is equal to the integral from 0 to infinity let me let me expand this out of e to the minus s X minus s C times f of X DX now what is this equal to well we could factor out an e to the minus SC and bring it outside of the integral because this has nothing to do with what we're taking the integral with respect to so let's do that let me take let me take this guy out and make just just to not confuse you let me rewrite the whole thing zero to infinity I could rewrite this Eater Maz II actually let me write it this way I'll do what was already in green as e to the minus s x times e to the minus s see right common base so if I were to multiply these two I could just add the exponents which you would get that up there times f of X D of X times f of X D of X this is a constant term with respect to X so we can just factor it out we can just factor this thing out right there so then you get e to the minus s C times the integral from 0 to infinity of e to the minus s x times f of X DX now what what were we doing here the whole time we were taking the Laplace transform we were taking the Laplace transform of we were taking the Laplace transform of the unit step function that goes up to see and then it's 0 up to see it and it's one after that of T times times some shifted function f of T minus C and now we got that as being equal to this thing and we made a substitution we simplified it a little bit e to the minus s see e to the minus s C times the integral from 0 to infinity of e to the minus s X f of X DX something about my the tablet doesn't work properly right around right around this period but this this this should look interesting to you what is this this this is the Laplace transform of f of X let me write that down what's the Laplace transform of well I could write it as f of T or f of X the Laplace transform of F of T is equal to the integral from 0 to infinity of e to the minus s T times f of T DT I mean the only and this are the exact same thing we're just using a tea here we're using an X here no difference they're just letters but they're lonely this is f of T this is f of T e to the minus s T times F of T DT I could have also very written is the Laplace transform of F of T I could write this as the integral from 0 to infinity of e to the minus s y times f of Y dy I could do it by anything because this is a definite integral the Y's you're going to disappear and we've seen that all you're left with is a function of s these this is all this ends up being some capital well you know we could write some capital function of s so this is interesting this is a Laplace transform of f of t times some scaling factor and that's what we set out to show so we can now show that the Laplace transform the Laplace transform of the unit step function times some function t minus C is equal to this function right here e to the minus s see where this C is the same as this C right here times the Laplace transform of f of t times the Laplace transform I don't know what's going on with the tablet right there of f of T let me write that this is equal to because it's looking funny there e to the minus s C times the Laplace transform of F of T so this is our result this is our result now what does this mean let's see if we know look at like backfilled it somehow what does this mean what can we what can we do with this well let's say we had the Laplace transform let's say we wanted to figure out the Laplace transform of the unit step function that starts up at PI of T and let's say we're taking I don't know something that we know well sine sine of sine of t minus pi so we shifted it right this thing is really malfunctioning at this point right here let me pause it I just Paulo sorry if that was a little disconcerting I just paused the video because it was having trouble recording at some point on my little board so let me rewrite the result that we'd proved just now we showed that the Laplace transform the Laplace transform of the unit step function T and it goes to one it's some value C times some function that's shifted by C to the right it's equal to e to the minus CS times the Laplace transform of just the unshifted function that was our result that was a big takeaway from this video and if this seems like some Byzantine hard to understand result we can apply it so let's say the Laplace transform is this what I was doing right before the actual pen tablet started started malfunctioning if we want to take the Laplace transform of the unit step function that goes to one at PI T times the sine function shifted by PI to the right we know that this is going to be equal to e to the minus CS C is PI in this case so minus PI s times the Laplace transform of the unshifted function so in this case it's the Laplace transform of sine of T and we know what the Laplace transform of sine of T is it's just 1 over s squared plus 1 so the plus transform of this thing here which before this video seemed like something crazy we now is we now know is this times this so it's e to the minus PI s times this or we can just write it as e to the minus PI s over s squared plus 1 we'll do a couple more examples of this in the next video where we go back and forth between the Laplace world and the the T the between the S domain and the time domain and I'll show you how this is a very useful result to take a lot of Laplace transforms and to invert a lot of Laplace transforms